Consider
$$\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{1-\cos^k\left({2\pi\over n}\right)}=L\tag1$$
How does one show that $L=\pi$ for $k>0$?
An attempt:
For $k=2$
$$\lim_{n\to \infty}{n\over 2}\cdot\sqrt{1-\cos^2\left({2\pi\over n}\right)}=L\tag2$$
$$\lim_{n\to \infty}{n\over 2}\cdot\sin\left({2\pi\over n}\right)=L\tag3$$
$$\lim_{n\to \infty}\pi\cdot{\sin\left({2\pi\over n}\right)\over {2\pi\over n}}=L\tag4$$
$$L=\pi\tag5$$
Hint. One may use a Taylor expansion of $\cos^kx$ as $x \to 0$, getting $$ \cos^kx=1-\frac{k}2x^2+O(x^4) $$ thus, as $x \to 0^+$, $$ \sqrt{1-\cos^kx}=\sqrt{\frac{k}2}x+O(x^2) $$ then putting $x=\dfrac{2\pi}n$, as $n \to \infty$.
HINT:
Assuming $k$ to be positive integer, try $$\dfrac{\left(1-\cos\dfrac{2\pi}n\right)}{\left(\dfrac\pi n\right)^2}\cdot\dfrac1k\sum_{r=0}^{k-1}\left(\cos\dfrac{2\pi}n\right)^r$$
With L'Hôpital (the expression makes sense for $n\not\in\Bbb N$): $$ \lim_{n\to\infty}{\frac{n^2}{2k}}\left(1-\cos^k\left(\frac{2\pi}n\right)\right) = \lim_{n\to\infty}\frac{1-\cos^k\left(\frac{2\pi}n\right)}{{\frac{2k}{n^2}}} = \lim_{n\to\infty}\frac{-\frac{2k\pi\sin\left(\frac{2\pi}n\right)\cos^{k-1}\left(\frac{2\pi}n\right)}{{n^2}}}{-\frac{4k}{n^3}} =\\ \lim_{n\to\infty}\frac12\pi n\sin\left(\frac{2\pi}n\right)\cos^{k-1}\left(\frac{2\pi}n\right) = \pi^2. $$ Now, take square root.
$$x\to 0 \\\cos x=1-\frac{1}2x^2+\frac{1}{4!}x^4+...\sim 1-\dfrac12x^2\\ \cos ^kx\sim (1-\frac{1}2x^2)^k \sim 1-\dfrac k2x^2\\ $$ now put into ,$\frac{2\pi}{n} \to 0$ $$\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{1-\cos^k\left({2\pi\over n}\right)}=\\ \lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{1-(1-\dfrac k2(\frac{2\pi}{n})^2)}\\= \lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{\dfrac k2(\frac{2\pi}{n})^2}\\= \lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{\dfrac k2}\times (\frac{2\pi}{n})=\\ \lim_{n\to \infty}{1\over \sqrt{2}}\cdot\sqrt{\dfrac 12}\times (\frac{2\pi}{1})=\dfrac{2\pi}{2}=\pi$$
